Problem 31 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=\tan ^{-1} x$$

Step-by-Step Solution

Answer

Based on the analysis and solution provided, the function \(f(x) = \tan^{-1}(x)\) is always increasing on its entire domain, which is \((-\infty, \infty)\). There are no decreasing intervals for this function.

Step 1: Find the derivative of the function

First, we will find the derivative of the given function \(f(x) = \tan^{-1}(x)\) using the chain rule. The derivative of \(\tan^{-1}(x)\) is already a well-known formula, which is: $$f'(x) = \frac{1}{1+x^2}$$ Now that we have the derivative, we can proceed to check the intervals where it is positive and negative.

Step 2: Determine the intervals for the derivative to be positive or negative

The derivative \(f'(x) = \frac{1}{1+x^2}\) is a fraction, and its numerator is always positive (1), and its denominator is always positive as well (since \(1+x^2 > 0\) for all x). Therefore, the entire fraction \(f'(x)\) is positive for all values of \(x\). Since the first derivative is positive for all x, the function \(f(x) = \tan^{-1}(x)\) is increasing on the entire domain.

Step 3: Write down the final increasing and decreasing intervals

The function \(f(x) = \tan^{-1}(x)\) has an increasing interval on its entire domain: - Increasing interval: \((-\infty, \infty)\) - Since the function is always increasing, there are no decreasing intervals.

Key Concepts

Derivative

Understanding derivatives is vital in calculus since they reveal how functions change. The derivative of a function measures its rate of change or the slope of the tangent line at any point. For the function given, \(f(x) = \tan^{-1}(x)\), its derivative is derived from a known calculus formula. By differentiating \(\tan^{-1}(x)\), you obtain:

• \(f'(x) = \frac{1}{1+x^2}\)

This derivative highlights important characteristics about the function's behavior. Since the formula \(\frac{1}{1+x^2}\) involves a fraction, let’s break down why it’s crucial:

• The numerator is 1, a constant value.
• The denominator, \(1+x^2\), is positive for all real numbers \(x\) because squares of numbers are non-negative.

Thus, the derivative \(f'(x)\) is positive over the entire domain, which signifies a monotonically increasing function.

Chain Rule

The chain rule is a powerful tool in calculus used for finding derivatives of composite functions. A composite function is one where a function is applied inside another function, such as \(h(x) = g(f(x))\). To differentiate such functions, the chain rule provides a structured approach:

• Identify the outer function \(g(u)\) and the inner function \(f(x)\).
• Differentiate the outer function \(g(u)\) with respect to the inner function \(u\).
• Differentiated the inner function \(f(x)\) with respect to \(x\).

This exercise used the chain rule implicitly, even though \(\tan^{-1}(x)\) is straightforward. Learning to recognize when to apply it is crucial for more complex derivatives.

Increasing and Decreasing Intervals

In calculus, identifying where a function increases or decreases is essential for graphing and understanding function behavior. To determine these intervals, we inspect the derivative:

• If the derivative \(f'(x) > 0\), the function is increasing on that interval.
• If the derivative \(f'(x) < 0\), the function is decreasing on that interval.

For \(f(x) = \tan^{-1}(x)\), since \(f'(x) = \frac{1}{1+x^2}\) is positive for all \(x\), the function is continuously increasing on its entire domain. Therefore, it is increasing for:

• Interval: \((-\infty, \infty)\)

Because the derivative never changes sign (it remains positive), there are no intervals where the function is decreasing.